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anonymous
 4 years ago
change the cartesian integral into an equivalent polar integral:
anonymous
 4 years ago
change the cartesian integral into an equivalent polar integral:

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{\sqrt{2}}^{2}\int\limits_{\sqrt{4y^{2}}}^{y} dx dy\]

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1dw:1353881320734:dwi think this is what it looks like at least

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1remember that x=rcos(t) y=rsin(t) and r=sqrt(x^2+y^2), but thats really already given as 2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think I understand how to get pi/4 < theta < pi/2 But finding the bounds for r is throwing me off.

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1well, since r only really changes from 0 to 2; the bounds for r are 0 to 2

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1it r changed according to some function of t the r would be bound from 0 to f(t)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0according to the answers, r is bounded by 2 < r < 2csc(theta)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1dw:1353881730467:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but at pi/2 r can be 2 and rad(2)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1it might help to see the limits in this fashion to be able to see how to change them out \[\Large \int_{y=\sqrt{2}}^{y=2}~~~\int_{x=\sqrt{4y^{2}}}^{x=y} ~~~dx~dy\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0my book says to plug in x=rcos(theta) etc and magically gets the polar bounds. They dont show any intermediate steps so im really confused.

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1\[x= \sqrt{r^2y^2}\]\[2cos(t) = \sqrt{2^2(2sin(t))^2}\]\[4cos^2(t) = 44sin^2(t)\]\[cos^2(t) +sin^2(t)=1~\text{for any t}\] \[x= y\]\[2cos(t) = 2sin(t)\]\[t=\frac{pi}4\] something along those lines i believe

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1y = 2 2sin(t) = 2 sin(t) = 1; when t=pi/2 y=sqrt(2) 2sin(t)=sqrt(2) sin(t)=sqrt(2)/2; when t=pi/4

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1i might need to see the steps in your books to refresh me memory

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0this is all that they show.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I just dont understand where they cot the csc from.

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1your d parts in that png are dydx ... not dxdy

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well, thats a problem... the question has it as dx dy

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I dont think dy dx even works. Wouldnt it be y=y y=rad(4y^2)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1yeah, so the png might be a typo

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok so ignoring the png, would the r just be 0 to 2?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1from what i can tell yes; dw:1353883154486:dw

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1but then i might have been thinking of this backwards

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1so, r goes from 2 up to the line; its not the area inside the circle

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1if i were to attempt this; I would determine the area of the triangle and subtract out the area inside the circle part

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0But doesnt the outer circle ask that the Y bounds are 2 and rad(2) not the X?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1i positioned the graph in a convenient manner; a swaped the x and y axis ,,, yes

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1so, what we want is a function in ploar from for the line y=2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well, even though the answer key messed up the dxdy, but 2(pi/2) seems to be the right answer.

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1it is; and youd be able to see that y=2 converts to 2csc(t) if i could remember how to do it :)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1got it r^2 = x^2 + y^2 ; but y=2 is constant r^2 = x^2 + 4 ; x = rcos(t) r^2 = r^2cos^2(t) + 4 r^2r^2cos^2(t) = 4 r^2(1cos^2(t)) = 4 r^2 sin^2(t) = 4 r^2 = 4csc^2(t) r = 2csc(t)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1area of triangle = 2*2/2 = 2 area of sector = 4pi/8 = pi/2 area of shaded region = triangle  sector: 2  pi/2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok that makes sense.. one more question, is it a coincidence that y = 2 r sin(t) = 2 r = 2 / sin(t) using 1/sin(t) = csc(t) trig identity r = 2 csc(t) ?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1not sure if its coincidence or not; but that seems to work out as well :)
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